## MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

Question 1.

Find the area of each of the following parallelograms:

Solution:

Area of parallelogram = Base × Height

(a) Height = 4 cm, Base = 7 cm

Area of parallelogram = 7 × 4 = 28 cm^{2}

(b) Height = 3 cm, Base = 5 cm

Area of parallelogram = 5 × 3 = 15 cm^{2}

(c) Height = 3.5 cm, Base = 2.5 cm

Area of parallelogram = 2.5 × 3.5 = 8.75 cm^{2}

(d) Height = 4.8 cm, Base = 5 cm

Area of parallelogram = 5 × 4.8 = 24 cm^{2}

(e) Height = 4.4 cm, Base = 2 cm

Area of parallelogram = 2 × 4.4 = 8.8 cm^{2}

Question 2.

Find the area of each of the following triangles:

Solution:

Area of triangle = \(\frac{1}{2}\) × Base × Height

(a) Base = 4 cm, height = 3 cm

Area = \(\frac{1}{2}\) × 4 × 3 = 6 cm^{2}

(b) Base = 5 cm, height = 3.2 cm

Area = \(\frac{1}{2}\) × 5 × 3.2 = 8 cm^{2}

(c) Base = 3 cm, height = 4 cm

Area = \(\frac{1}{2}\) × 3 × 4 = 6cm^{2}

(d) Base = 3 cm, height = 2 cm

Area = \(\frac{1}{2}\) × 3 × 2 = 3 cm^{2}

Question 3.

Find the missing values:

Solution:

Area of parallelogram = Base × Height

(a) Base = 20 cm

Let height = h

Area of parallelogram = 246 cm^{2}

∴ 20 × h = 246

⇒ h = \(\frac{246}{20}\) = 12.3 cm

Therefore, the height of parallelogram is 12.3 cm.

(b) Let base = b

Height = 15 cm

Area of parallelogram = 154.5 cm^{2}

∴ b × 15 = 154.5

⇒ b = \(\frac{154.5}{15}\) = 10.3 cm

Therefore, the base of parallelogram is 10.3 cm.

(c) Let base = b

Height = 8.4 cm

Area of parallelogram = 48.72 cm^{2}

∴ b × 8.4 = 48.72

⇒ b = \(\frac{48.72}{8.4}\) = 5.8 cm

Therefore, the base of parallelogram is 5.8 cm.

(d) Base = 15.6 cm

Let height = h

Area of parallelogram = 16.38 cm^{2}

∴15.6 × h = 16.38

⇒ h = \(\frac{16.38}{15.6}\) = 1.05 cm

Therefore, the height of parallelogram is 1.05 cm.

Question 4.

Find the missing values:

Solution:

Area of triangle = \(\frac{1}{2}\) × Base × Height

Let b be the base of triangle and h be the height of triangle.

(i) b = 15 cm

Therefore, the height of triangle is 11.6 cm.

(ii) h = 31.4 mm

Area = \(\frac{1}{2}\) × b × h = 1256 mm^{2}

Therefore, the base of triangle is 80 mm.

(iii) b = 22 cm

Therefore, the height of triangle is 15.5 cm.

Question 5.

PQRS is a parallelogram (see the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:

(a) the area of the parallelogram PQRS

(b) QN, if PS = 8 cm

Solution:

(a) Area of parallelogram = Base × Height

= SR × QM

= 12 × 7.6 = 91.2 cm^{2}

(b) PS = 8 cm

Area of parallelogram = Base × Height

= PS × QN = 91.2 cm^{2}

⇒ 8 × QN = 91.2

⇒ QN = \(\frac{91.2}{8}\) = 11.4 cm

Question 6.

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (see the given figure).

If the area of the parallelogram is 1470 cm^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:

Area of parallelogram = Base × Height

= AB × DL

⇒ 1470 = 35 × DL

⇒ DL = \(\frac{1470}{35}\) = 42 cm

Also, area of parallelogram = AD × BM

⇒ 1470 = 49 × BM

∴ BM = \(\frac{1470}{49}\) = 30 cm

Question 7.

∆ABC is right angled at A (see the given figure). AD is perpendicular to BC. If AB = 5 cm, BC – 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.

Solution:

Question 8.

∆ABC is isosceles with AB = AC= 7.5 cm and BC = 9 cm (see the given figure). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i. e., CE?

Solution: